Coin Problem 6 Solution

This is the coin problem 6 solution which read "a stack of pennies and dimes has a total value of $2.31. how many dimes are in the stack if there are twice as many dimes as pennies?"

make $2.31 into cents so 231 cents .. a dime is worth 10x and penny is worth x.  So if you had 1 dime and 1 penny you would have 10(1) or 10 cents plus 1(1) or 1 cent which is 11 cents.  Were trying to have it add up to 231 cents though and there are twice as many dimes 2(10x) or 20x as pennies.  

So we have 20x + x = 231


21x = 231

x = 11  So there are 11 penny's or 11 cents and 22 dimes or 220 cents .. 220 cents plus 11 cents equals 231 cents or $2.31.

Coin Problem 5 Solution

Let the total amount in cents = 2251

let x = the # pennies and the value = 1(x) or just x

let 2x = the # of quarters and 25(2x) or 50x = the value in quarters

let 20(2x) or 40x equal the # of 50 cent pieces and 50(40x) or 2000x = the value of 50 cent pieces

So, we have:

x + 50x + 2000x = 2051
2051x = 2051
x = 2051/2051
x = 1

so we have (1) pennies

2(1) or 2 quarters

and 40(1) or 40 half dollars

Coin Problem 4 Solution

19.43 is 1943 cents.

In this problem, it helps to think about what the easiest setup would be before just setting x equal to the 1st thing you see. If you let x = pennies, well that would work, but you would end up worth a lot of fractions that you could avoid if you started with x = quarters instead.

So let x = the # of quarters and 25x the value

2x = the number of dimes and 10(2x) = the value or 20x

4x = the # of pennies and 1(4x) or 4x the value of pennies

4x + 1 = the # nickels and 5(4x+1) or 20x + 5 the value of nickels

and

4(4x+1) or 16x + 4 the # of half dollars and 50(16x + 4) or 800x + 200 the value

So altogether we have 25x + 20x + 4x + 20x + 5 + 800x + 200 = 1943

869x + 205 = 1943

869x = 1738

x = 1738/869

x = 2

So we have 2 quarters, (50 cents)

2(2) or 4 dimes (40 cents)

4(2) or 8 pennies (8 cents)

4(2) + 1 or 9 nickels (45 cents)

and 16(2) + 4 or 36 half dollars (1800 cents)

just to check if you add those cents up it will equal 1943 cents which is $19.43

Coin Problem 3 Solution

Again make $10.35 equal to cents --- 1035 cents 1st.

Algebraically we have

x = 32x 32 cent stamps

x + 2 = 25(x + 2) 25 cent stamps and

2(x + 2) or 2x + 4 = 50(2x + 4) 50 cent stamps

So, we then have..

32x + 25(x + 2) + 50(2x + 4) = 1035

32x + 25x + 50 + 100x + 200 = 1035

157x + 250 = 1035

157x = 1035 - 250

157x = 785

x = 785/157

x = 5

So we have 5 32 cent stamps, 5 + 2 or 7 25 cent stamps, and 2(5 + 2) or 14 50 cent stamps...

Coin Problem 2 Solution

1st of all on all these problems convert the total amount of money from dollar to cents-- $100 is equal to 10000 cents (just add 2 zeros), $7 is equal to 700 cents, and you already have 37 cents, so altogether you have 10000 + 600 + 87 = 10737 cents.

From the problem you can see that everything is relative to pennies so let that be 1(x) or just x

1 more than twice as many nickels as pennies is (2x + 1) and and since its nickels 5(2x +1) or 10x + 5

twice as many dimes as nickels 2(2x +1) = 4x + 2 and since its dimes 10(4x + 2) or 40x + 20

1 more than twice as many 50 cents pieces than dimes (this is tricky wording since you may expect the denomination to go to a quarter) which is 2(4x + 2) + 1 or 8x + 5 and since its 50 cent pieces 50(8x + 5) or 400x + 250

and twice as many quarters as 50 cent pieces or 2(8x + 5) or 16x + 10 and since its quarters 25(16x + 10) or 400x + 250

so algebraically we have
x + 10x + 5 + 40x + 20 + 400x + 250 + 400x + 250 = 10737 =
851x + 525 = 10737 =
851x = 10737 - 525 =
851x = 10212 =
x = 10212/851

x = 12

So you have 12 pennies, 2(12) + 1 or 25 nickels, 4(12) + 2 or 50 dimes, 8(12) + 5 or 101 Fifty cent pieces, and 16(12) + 10 or 202 quarters....

Yeah, a little tedious arithmetic wise...

Coin Problem 1 Solution

On any kind of coin problem its important to turn any dollar amount (if given in dollars) to cents and to make sure you give the appropriate value of each denomination ---- a penny 1 or -- nickel 5--- dimes 10-- quarter 25-- sounds kind of obvious but a lot of problems are taken care of if you just know that information for sure.

Alright if you have 10 pennies you have 10(1) == 10 cents So,

If you have 12 nickels you would have 12(5) = 60 cents

If you had 5 dimes you would have 5(10) = 50 cents

and if you had 10 quarters you would have 10(25) = 250 cents or $2.50

To solve this problem you 1st set up the information given. You can see that everything is relative to the pennies (1 more nickel than pennies and 6 times as many dimes as pennies)
So let pennies equal 1x or just x since 1 times anything is just the number...

you have x + 1 nickels at 5 cents so 5(x+1) nickels

and 6 times as many dimes as pennies or 10(6x) or 60x

Theres no need to change the dollars to cents as the cents are already given 71 cents.

So the equation would be

x + 5(x+1) + 10(6x) = 71 =

x + 5x + 5 + 60x = 71 =

66x = 66 =
x = 66/66 = 1

So thers 1 penny, (1 + 1) or 2 nickels and 6(1) 6 dimes

To check if you add those up 1 + 10 + 60 you get 71 so the answer is correct.